NOTE: This ap-calculus EDG will be closing in the next few weeks. Please sign up for the new AP CalculusTeacher Community Forum at https://apcommunity.collegeboard.org/getting-startedand post messages there. ------------------------------------------------------------------------------------------------I would argue that y=x^(1/3) has a point of inflection at the origin, even thought the function is not differentiable there. While a function that changes concavity via a "corner" does not have a point of inflection at said point. "My" defn of an inflection point is a "smooth change in concavity". This allows for vertical tangents, but not cusps and certainly not points of discontinuity. Of course, then my students ask what I mean by "smooth", which is a good thing!

Bradley

"...each day's a gift and not a given rightLeave no stone unturned, leave your fears behindAnd try to take the path less traveled byThat first step you take is the longest stride."

NOTE:This ap-calculus EDG will be closing in the next few weeks. Please sign up for the new AP CalculusTeacher Community Forum at https://apcommunity.collegeboard.org/getting-startedand post messages there.------------------------------------------------------------------------------------------------I find conflicting reports on this, which leads me to believe there may be conflicting opinions or varying explanations among textbooks. For that reason, I assume this question would not be addressed in this way on the exam.

Can a point of inflection be identified where the function has a vertical asymptote just because the concavity changes? For example does y=1/x have a point of inflection at x=0? My belief is that a point of inflection cannot exist at a point where the function is not defined or even not differentiable.